3.232 \(\int \sqrt {x} \tanh ^{-1}(\tanh (a+b x))^{5/2} \, dx\)

Optimal. Leaf size=174 \[ -\frac {5 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))}}\right ) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4}{64 b^{3/2}}+\frac {5}{32} x^{3/2} \sqrt {\tanh ^{-1}(\tanh (a+b x))} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2-\frac {5}{24} x^{3/2} \tanh ^{-1}(\tanh (a+b x))^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )+\frac {1}{4} x^{3/2} \tanh ^{-1}(\tanh (a+b x))^{5/2}-\frac {5 \sqrt {x} \sqrt {\tanh ^{-1}(\tanh (a+b x))} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}{64 b} \]

[Out]

-5/64*arctanh(b^(1/2)*x^(1/2)/arctanh(tanh(b*x+a))^(1/2))*(b*x-arctanh(tanh(b*x+a)))^4/b^(3/2)-5/24*x^(3/2)*(b
*x-arctanh(tanh(b*x+a)))*arctanh(tanh(b*x+a))^(3/2)+1/4*x^(3/2)*arctanh(tanh(b*x+a))^(5/2)+5/32*x^(3/2)*(b*x-a
rctanh(tanh(b*x+a)))^2*arctanh(tanh(b*x+a))^(1/2)-5/64*(b*x-arctanh(tanh(b*x+a)))^3*x^(1/2)*arctanh(tanh(b*x+a
))^(1/2)/b

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Rubi [A]  time = 0.10, antiderivative size = 174, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2169, 2165} \[ -\frac {5 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))}}\right ) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4}{64 b^{3/2}}+\frac {5}{32} x^{3/2} \sqrt {\tanh ^{-1}(\tanh (a+b x))} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2-\frac {5}{24} x^{3/2} \tanh ^{-1}(\tanh (a+b x))^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )+\frac {1}{4} x^{3/2} \tanh ^{-1}(\tanh (a+b x))^{5/2}-\frac {5 \sqrt {x} \sqrt {\tanh ^{-1}(\tanh (a+b x))} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}{64 b} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[x]*ArcTanh[Tanh[a + b*x]]^(5/2),x]

[Out]

(-5*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[ArcTanh[Tanh[a + b*x]]]]*(b*x - ArcTanh[Tanh[a + b*x]])^4)/(64*b^(3/2)) + (
5*x^(3/2)*(b*x - ArcTanh[Tanh[a + b*x]])^2*Sqrt[ArcTanh[Tanh[a + b*x]]])/32 - (5*Sqrt[x]*(b*x - ArcTanh[Tanh[a
+ b*x]])^3*Sqrt[ArcTanh[Tanh[a + b*x]]])/(64*b) - (5*x^(3/2)*(b*x - ArcTanh[Tanh[a + b*x]])*ArcTanh[Tanh[a +
b*x]]^(3/2))/24 + (x^(3/2)*ArcTanh[Tanh[a + b*x]]^(5/2))/4

Rule 2165

Int[1/(Sqrt[u_]*Sqrt[v_]), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[(2*ArcTanh[(
Rt[a*b, 2]*Sqrt[u])/(a*Sqrt[v])])/Rt[a*b, 2], x] /; NeQ[b*u - a*v, 0] && PosQ[a*b]] /; PiecewiseLinearQ[u, v,
x]

Rule 2169

Int[(u_)^(m_)*(v_)^(n_.), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[(u^(m + 1)*v^
n)/(a*(m + n + 1)), x] - Dist[(n*(b*u - a*v))/(a*(m + n + 1)), Int[u^m*v^(n - 1), x], x] /; NeQ[b*u - a*v, 0]]
/; PiecewiseLinearQ[u, v, x] && NeQ[m + n + 2, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !In
tegerQ[n] || LtQ[0, m, n])) &&  !ILtQ[m + n, -2]

Rubi steps

\begin {align*} \int \sqrt {x} \tanh ^{-1}(\tanh (a+b x))^{5/2} \, dx &=\frac {1}{4} x^{3/2} \tanh ^{-1}(\tanh (a+b x))^{5/2}-\frac {1}{8} \left (5 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )\right ) \int \sqrt {x} \tanh ^{-1}(\tanh (a+b x))^{3/2} \, dx\\ &=-\frac {5}{24} x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{3/2}+\frac {1}{4} x^{3/2} \tanh ^{-1}(\tanh (a+b x))^{5/2}+\frac {1}{16} \left (5 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2\right ) \int \sqrt {x} \sqrt {\tanh ^{-1}(\tanh (a+b x))} \, dx\\ &=\frac {5}{32} x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}-\frac {5}{24} x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{3/2}+\frac {1}{4} x^{3/2} \tanh ^{-1}(\tanh (a+b x))^{5/2}-\frac {1}{64} \left (5 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3\right ) \int \frac {\sqrt {x}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))}} \, dx\\ &=\frac {5}{32} x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}-\frac {5 \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{64 b}-\frac {5}{24} x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{3/2}+\frac {1}{4} x^{3/2} \tanh ^{-1}(\tanh (a+b x))^{5/2}+\frac {\left (5 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )\right ) \int \frac {1}{\sqrt {x} \sqrt {\tanh ^{-1}(\tanh (a+b x))}} \, dx}{128 b}\\ &=-\frac {5 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))}}\right ) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4}{64 b^{3/2}}+\frac {5}{32} x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}-\frac {5 \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{64 b}-\frac {5}{24} x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{3/2}+\frac {1}{4} x^{3/2} \tanh ^{-1}(\tanh (a+b x))^{5/2}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 121, normalized size = 0.70 \[ \frac {\sqrt {x} \sqrt {\tanh ^{-1}(\tanh (a+b x))} \left (-55 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))+73 b x \tanh ^{-1}(\tanh (a+b x))^2+15 \tanh ^{-1}(\tanh (a+b x))^3+15 b^3 x^3\right )}{192 b}-\frac {5 \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^4 \log \left (\sqrt {b} \sqrt {\tanh ^{-1}(\tanh (a+b x))}+b \sqrt {x}\right )}{64 b^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[x]*ArcTanh[Tanh[a + b*x]]^(5/2),x]

[Out]

(Sqrt[x]*Sqrt[ArcTanh[Tanh[a + b*x]]]*(15*b^3*x^3 - 55*b^2*x^2*ArcTanh[Tanh[a + b*x]] + 73*b*x*ArcTanh[Tanh[a
+ b*x]]^2 + 15*ArcTanh[Tanh[a + b*x]]^3))/(192*b) - (5*(-(b*x) + ArcTanh[Tanh[a + b*x]])^4*Log[b*Sqrt[x] + Sqr
t[b]*Sqrt[ArcTanh[Tanh[a + b*x]]]])/(64*b^(3/2))

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fricas [A]  time = 0.40, size = 162, normalized size = 0.93 \[ \left [\frac {15 \, a^{4} \sqrt {b} \log \left (2 \, b x - 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) + 2 \, {\left (48 \, b^{4} x^{3} + 136 \, a b^{3} x^{2} + 118 \, a^{2} b^{2} x + 15 \, a^{3} b\right )} \sqrt {b x + a} \sqrt {x}}{384 \, b^{2}}, \frac {15 \, a^{4} \sqrt {-b} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right ) + {\left (48 \, b^{4} x^{3} + 136 \, a b^{3} x^{2} + 118 \, a^{2} b^{2} x + 15 \, a^{3} b\right )} \sqrt {b x + a} \sqrt {x}}{192 \, b^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(1/2)*arctanh(tanh(b*x+a))^(5/2),x, algorithm="fricas")

[Out]

[1/384*(15*a^4*sqrt(b)*log(2*b*x - 2*sqrt(b*x + a)*sqrt(b)*sqrt(x) + a) + 2*(48*b^4*x^3 + 136*a*b^3*x^2 + 118*
a^2*b^2*x + 15*a^3*b)*sqrt(b*x + a)*sqrt(x))/b^2, 1/192*(15*a^4*sqrt(-b)*arctan(sqrt(b*x + a)*sqrt(-b)/(b*sqrt
(x))) + (48*b^4*x^3 + 136*a*b^3*x^2 + 118*a^2*b^2*x + 15*a^3*b)*sqrt(b*x + a)*sqrt(x))/b^2]

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giac [A]  time = 0.17, size = 204, normalized size = 1.17 \[ \frac {1}{384} \, \sqrt {2} {\left (48 \, \sqrt {2} {\left (\sqrt {b x + a} {\left (2 \, x + \frac {a}{b}\right )} \sqrt {x} + \frac {a^{2} \log \left ({\left | -\sqrt {b} \sqrt {x} + \sqrt {b x + a} \right |}\right )}{b^{\frac {3}{2}}}\right )} a^{2} + 16 \, \sqrt {2} {\left (\sqrt {b x + a} {\left (2 \, {\left (4 \, x + \frac {a}{b}\right )} x - \frac {3 \, a^{2}}{b^{2}}\right )} \sqrt {x} - \frac {3 \, a^{3} \log \left ({\left | -\sqrt {b} \sqrt {x} + \sqrt {b x + a} \right |}\right )}{b^{\frac {5}{2}}}\right )} a b + \sqrt {2} {\left ({\left (2 \, {\left (4 \, {\left (6 \, x + \frac {a}{b}\right )} x - \frac {5 \, a^{2}}{b^{2}}\right )} x + \frac {15 \, a^{3}}{b^{3}}\right )} \sqrt {b x + a} \sqrt {x} + \frac {15 \, a^{4} \log \left ({\left | -\sqrt {b} \sqrt {x} + \sqrt {b x + a} \right |}\right )}{b^{\frac {7}{2}}}\right )} b^{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(1/2)*arctanh(tanh(b*x+a))^(5/2),x, algorithm="giac")

[Out]

1/384*sqrt(2)*(48*sqrt(2)*(sqrt(b*x + a)*(2*x + a/b)*sqrt(x) + a^2*log(abs(-sqrt(b)*sqrt(x) + sqrt(b*x + a)))/
b^(3/2))*a^2 + 16*sqrt(2)*(sqrt(b*x + a)*(2*(4*x + a/b)*x - 3*a^2/b^2)*sqrt(x) - 3*a^3*log(abs(-sqrt(b)*sqrt(x
) + sqrt(b*x + a)))/b^(5/2))*a*b + sqrt(2)*((2*(4*(6*x + a/b)*x - 5*a^2/b^2)*x + 15*a^3/b^3)*sqrt(b*x + a)*sqr
t(x) + 15*a^4*log(abs(-sqrt(b)*sqrt(x) + sqrt(b*x + a)))/b^(7/2))*b^2)

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maple [B]  time = 0.24, size = 471, normalized size = 2.71 \[ \frac {\sqrt {x}\, \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {7}{2}}}{4 b}-\frac {a \sqrt {x}\, \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {5}{2}}}{24 b}-\frac {5 a^{2} \sqrt {x}\, \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{96 b}-\frac {5 a^{3} \sqrt {x}\, \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{64 b}-\frac {5 \ln \left (\sqrt {b}\, \sqrt {x}+\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}\right ) a^{4}}{64 b^{\frac {3}{2}}}-\frac {5 a^{3} \ln \left (\sqrt {b}\, \sqrt {x}+\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}\right ) \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )}{16 b^{\frac {3}{2}}}-\frac {15 a^{2} \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right ) \sqrt {x}\, \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{64 b}-\frac {15 a^{2} \ln \left (\sqrt {b}\, \sqrt {x}+\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}\right ) \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}}{32 b^{\frac {3}{2}}}-\frac {5 a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right ) \sqrt {x}\, \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{48 b}-\frac {15 a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2} \sqrt {x}\, \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{64 b}-\frac {5 a \ln \left (\sqrt {b}\, \sqrt {x}+\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}\right ) \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{3}}{16 b^{\frac {3}{2}}}-\frac {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right ) \sqrt {x}\, \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {5}{2}}}{24 b}-\frac {5 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2} \sqrt {x}\, \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{96 b}-\frac {5 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{3} \sqrt {x}\, \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{64 b}-\frac {5 \ln \left (\sqrt {b}\, \sqrt {x}+\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}\right ) \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{4}}{64 b^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(1/2)*arctanh(tanh(b*x+a))^(5/2),x)

[Out]

1/4*x^(1/2)*arctanh(tanh(b*x+a))^(7/2)/b-1/24/b*a*x^(1/2)*arctanh(tanh(b*x+a))^(5/2)-5/96/b*a^2*x^(1/2)*arctan
h(tanh(b*x+a))^(3/2)-5/64/b*a^3*x^(1/2)*arctanh(tanh(b*x+a))^(1/2)-5/64/b^(3/2)*ln(b^(1/2)*x^(1/2)+arctanh(tan
h(b*x+a))^(1/2))*a^4-5/16/b^(3/2)*a^3*ln(b^(1/2)*x^(1/2)+arctanh(tanh(b*x+a))^(1/2))*(arctanh(tanh(b*x+a))-b*x
-a)-15/64/b*a^2*(arctanh(tanh(b*x+a))-b*x-a)*x^(1/2)*arctanh(tanh(b*x+a))^(1/2)-15/32/b^(3/2)*a^2*ln(b^(1/2)*x
^(1/2)+arctanh(tanh(b*x+a))^(1/2))*(arctanh(tanh(b*x+a))-b*x-a)^2-5/48/b*a*(arctanh(tanh(b*x+a))-b*x-a)*x^(1/2
)*arctanh(tanh(b*x+a))^(3/2)-15/64/b*a*(arctanh(tanh(b*x+a))-b*x-a)^2*x^(1/2)*arctanh(tanh(b*x+a))^(1/2)-5/16/
b^(3/2)*a*ln(b^(1/2)*x^(1/2)+arctanh(tanh(b*x+a))^(1/2))*(arctanh(tanh(b*x+a))-b*x-a)^3-1/24/b*(arctanh(tanh(b
*x+a))-b*x-a)*x^(1/2)*arctanh(tanh(b*x+a))^(5/2)-5/96/b*(arctanh(tanh(b*x+a))-b*x-a)^2*x^(1/2)*arctanh(tanh(b*
x+a))^(3/2)-5/64/b*(arctanh(tanh(b*x+a))-b*x-a)^3*x^(1/2)*arctanh(tanh(b*x+a))^(1/2)-5/64/b^(3/2)*ln(b^(1/2)*x
^(1/2)+arctanh(tanh(b*x+a))^(1/2))*(arctanh(tanh(b*x+a))-b*x-a)^4

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {x} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{\frac {5}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(1/2)*arctanh(tanh(b*x+a))^(5/2),x, algorithm="maxima")

[Out]

integrate(sqrt(x)*arctanh(tanh(b*x + a))^(5/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \sqrt {x}\,{\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(1/2)*atanh(tanh(a + b*x))^(5/2),x)

[Out]

int(x^(1/2)*atanh(tanh(a + b*x))^(5/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(1/2)*atanh(tanh(b*x+a))**(5/2),x)

[Out]

Timed out

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